Chapter 3.1, Problem 42E

To determine

To find : the equation of line tangent to the curve at the point $t=-1/6$ .

Answer to Problem 42E

The equation of line tangent to the curve at point $t=-1/6$ is $y-\frac{1}{2}=\sqrt{3}\left(x+\frac{\sqrt{3}}{2}\right)$ .

Explanation of Solution

Given information :

  $x=\sin \left(2\pi t\right),y=\cos \left(2\pi t\right)$ .

Formula needed :

The equation of line tangent to the curve at point $t=a$ is $y-y_1=m\left(x-x_1\right)$ where $x_1,y_1$ are the value of $x$ and $y$ at $t=a$ respectively. Slope of line is $m={\left.\frac{dy}{dx}\right|}_{t=a}$ .

  $\frac{dy}{dx}$ partially expressed as $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ .

Partially differentiate $x$ and $y$ with respect to $t$ .

  $\begin{array}{c}\frac{dx}{dt}=\frac{d}{dt}\left(\sin \left(2\pi t\right)\right)\\ =2\pi \cos \left(2\pi t\right)\\ \frac{dy}{dt}=\frac{d}{dt}\left(\cos \left(2\pi t\right)\right)\\ =-2\pi \sin \left(2\pi t\right)\end{array}$

So,

  $\begin{array}{c}\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\\ =\frac{-2\pi \sin \left(2\pi t\right)}{2\pi \cos \left(2\pi t\right)}\\ =-\tan \left(2\pi t\right)\end{array}$

Slope is given by $m={\left.\frac{dy}{dx}\right|}_{t=-1/6}$ .

  $\begin{array}{c}m={\left.\frac{dy}{dx}\right|}_{t=-1/6}\\ =-\tan \left(2\pi \left(\frac{-1}{6}\right)\right)\\ =-\tan \left(\frac{-\pi }{3}\right)\\ =\sqrt{3}\end{array}$

The value of $x$ and $y$ at $t=-1/6$ is given by $x_1=\sin \left(-\frac{2\pi }{6}\right),y_1=\cos \left(-\frac{2\pi }{6}\right)$ .

  $\begin{array}{c}{x}_1=-\sin \left(\frac{\pi }{3}\right)\\ =-\frac{\sqrt{3}}{2}\\ y_1=\cos \left(\frac{\pi }{3}\right)\\ =\frac{1}{2}\end{array}$

Thus, the equation of line tangent to the curve at point $t=-1/6$ is $y-\frac{1}{2}=\sqrt{3}\left(x+\frac{\sqrt{3}}{2}\right)$ .